This time, we apply the inclusion-exclusion principle:
A fair die was rolled n times. What is the probability that at least 1 of the 6 values never appears?
The required probability is nothing but:
the probability of not getting a 1 OR
the probability of not getting a 2 OR
the probability of not getting a 3 OR
the probability of not getting a 4 OR
the probability of not getting a 5 OR
the probability of not getting a 6
The probabilities are the same since it is a fair die (5/6).
Let’s apply the inclusion-exclusion principle:
1) Add all single probabilities: The probability of a given number missing in one roll = 5/6. All such single probabilities in n rolls = 6C1 x (5/6)n
2) Subtract all pair probabilities: The probability of the given two numbers missing in one roll = 4/6. All such pair probabilities = 6C2 x (4/6)n
3) Add all 3-tuple probabilities: The probability of the given three numbers missing in one roll = 3/6. All such pair probabilities = 6C3 x (3/6)n
4) Subtract all 4-tuple probabilities: The probability of the given four numbers missing in one roll = 2/6. All such pair probabilities = 6C4 x (2/6)n
5) Add all 5-tuple probabilities: The probability of the given five numbers missing in one roll = 1/6. All such pair probabilities = 6C5 x (1/6)n
6C1 x (5/6)n – 6C2 x (4/6)n + 6C3 x (3/6)n – 6C4 x (2/6)n + 6C5 x (1/6)n
For ten rolls, n = 10
n <- 10
choose(6,1) * (5/6)^n - choose(6,2) * (4/6)^n + choose(6, 3)* (3/6)^n - choose(6,4)*(2/6)^n + choose(6,5) * (1/6)^n0.728Similarly, for 20 rolls, there is a 0.15 chance of not getting at least one number.
